Mathematics - Definite Integration Question with Solution | TestHub

$f(x)={\int }_0^x\frac{e^t}{t}dt$

$f(x+a)-f(1+a)={\int }_0^{a+x}\frac{e^{t}dt}{t}-{\int }_0^{a+1}\frac{e^t}{t}dt$

$f(x+a)-f(1+a)={\int }_{a+1}^{a+x}\frac{e^t}{t}dt$

Let $t+a=t^{\prime }\Rightarrow f(x+a)-f(1+a)={\int }_{a+1}^{a+x}\frac{e^{t^{\prime }}.e^a}{t^{\prime }+a}d{t}^{\prime }$

$f(x+a)-f(1+a)=e^a{\int }_1^x\frac{e^t}{t+a}dt$

$f(x+a)-f(1+a)=e^a{\int }_0^{x-1}\frac{e^{t-1}}{t+a-1}dt$